What is the force imparted by the bullet on the block

In summary, the bullet with a mass of 0.045 kg and a velocity of 560m/s forward collided with a stationary block with a mass of 0.5kg, resulting in a velocity of 46.2m/s forward for both objects. Using conservation of momentum, the average force exerted on the bullet by the block was calculated to be -66N forward, which is equal in magnitude but opposite in direction to the average force exerted on the block by the bullet. This is in accordance with Newton's 3rd law.
  • #1
hamza2095
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1

Homework Statement


A bullet with a mass of 0.045 kg is shot at 560m/s fwd, and it gets dislodged in a stationary block that is 0.5kg.
The collision lasts 0.35 seconds. What is the force imparted by the bullet on the block?

Homework Equations


F=(m(vf-vi))/t

The Attempt at a Solution


F=(0.045kg(0-560m/s))/0.35s
F=-72N [FWD]

It doesn't make sense that the force the bullet has on the block is backwards 72N[BWD], so I'm guessing that's the force of the block on the bullet, therefore using Newton's 3rd law which if F Action= -F reaction i get the force of the bullet on the block is 72N [FWD]

Is this right, shouldn't the initial calculation be the final answer for the question? Help is appreciated
 
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  • #2
hamza2095 said:

Homework Statement


A bullet with a mass of 0.045 kg is shot at 560m/s fwd, and it gets dislodged in a stationary block that is 0.5kg.
The collision lasts 0.35 seconds. What is the force imparted by the bullet on the block?

Homework Equations


F=(m(vf-vi))/t

The Attempt at a Solution


F=(0.045kg(0-560m/s))/0.35s
F=-72N [FWD]

It doesn't make sense that the force the bullet has on the block is backwards 72N[BWD], so I'm guessing that's the force of the block on the bullet, therefore using Newton's 3rd law which if F Action= -F reaction i get the force of the bullet on the block is 72N [FWD]

Is this right, shouldn't the initial calculation be the final answer for the question? Help is appreciated
Are you sure the "stationary block" remains stationary after the collision?

To me, the fact that the problem statement mentions the 0.5 kg mass of the block is a hint that the bullet+block combination has a non-zero velocity after the collision.
 
  • #3
collinsmark said:
Are you sure the "stationary block" remains stationary after the collision?

To me, the fact that the problem statement mentions the 0.5 kg mass of the block is a hint that the bullet+block combination has a non-zero velocity after the collision.
I get 46.2m/s (fwd) for the velocity of the block and bullet using conservation of momentum, and with that i still get -66N fwd. But I think you use the velocity of the bullet relative to the block, which is still 0 because it's inside of it.
 
  • #4
hamza2095 said:
I get 46.2m/s (fwd) for the velocity of the block and bullet using conservation of momentum, and with that i still get -66N fwd.

That sounds about right. :smile: The force applied to the bullet.

But I think you use the velocity of the bullet relative to the block, which is still 0 because it's inside of it.
You'll need to ensure that your frame of reference is the same before and after the collision.

You can't use a stationary frame before the collision, and then switch to a moving frame after, when calculating velocities and momentum values.
 
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  • #5
As an aside, the wording of the question could be better. It should make clear that the block is free to move, and it should not ask for "the force imparted". You can only calculate an average force, and forces are not imparted. Imparting implies transfer of some state, such as energy or momentum. Forces are applied or exerted.
 
  • #6
collinsmark said:
You'll need to ensure that your frame of reference is the same before and after the collision.

You can't use a stationary frame before the collision, and then switch to a moving frame after, when calculating velocities and momentum values.
Okay I see what your saying, so using conservation of momentum i get that both the block and bullet are going 46.2m/s FWD. If i use that final velocity to find force I get -66 N [FWD], which is still a negative.
 
  • #7
hamza2095 said:
Okay I see what your saying, so using conservation of momentum i get that both the block and bullet are going 46.2m/s FWD. If i use that final velocity to find force I get -66 N [FWD], which is still a negative.
You calculated the average force that slowed the bullet. That is therefore the force exerted on the bullet, but the question asks for the force the bullet exerted on the block.
 
  • #8
haruspex said:
As an aside, the wording of the question could be better. It should make clear that the block is free to move, and it should not ask for "the force imparted". You can only calculate an average force, and forces are not imparted. Imparting implies transfer of some state, such as energy or momentum. Forces are applied or exerted.
So according to Newton's 3rd law the average force the bullet exerted on the block should be 66N [FWD], correct?
 
  • #9
hamza2095 said:
So according to Newton's 3rd law the average force the bullet exerted on the block should be 66N [FWD], correct?
Yes.
 
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FAQ: What is the force imparted by the bullet on the block

1. What is the definition of force?

Force is a physical quantity that can be described as a push or pull on an object that causes it to accelerate or change its motion.

2. How is the force of a bullet on a block calculated?

The force of a bullet on a block can be calculated using the formula F=ma, where F is the force, m is the mass of the bullet, and a is the acceleration of the bullet upon impact.

3. Does the speed of the bullet affect the force it imparts on the block?

Yes, the speed of the bullet does affect the force it imparts on the block. The greater the speed, the greater the force of impact on the block.

4. What factors can influence the force of a bullet on a block?

The force of a bullet on a block can be influenced by factors such as the mass and velocity of the bullet, the material and density of the block, and the angle and type of impact.

5. How is the force of the bullet on the block related to Newton's Third Law of Motion?

According to Newton's Third Law of Motion, for every action, there is an equal and opposite reaction. This means that the force of the bullet on the block is equal and opposite to the force of the block on the bullet.

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